Question

ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Then angle BCD measures?

Answer 1

Answer:

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solution.

AB=AC         (Given)

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     Let ∠DBC=∠ACB=x         .......(1)

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     Let ∠DBC=∠ACB=x         .......(1)AC=AD          (Given)

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     Let ∠DBC=∠ACB=x         .......(1)AC=AD          (Given)It means that ∠ACD=∠BDC         (In triangle, angles opposite to equal sides are equal)     

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     Let ∠DBC=∠ACB=x         .......(1)AC=AD          (Given)It means that ∠ACD=∠BDC         (In triangle, angles opposite to equal sides are equal)     Let ∠ACD=∠BDC=y           ......(2)

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     Let ∠DBC=∠ACB=x         .......(1)AC=AD          (Given)It means that ∠ACD=∠BDC         (In triangle, angles opposite to equal sides are equal)     Let ∠ACD=∠BDC=y           ......(2)In ∆BDC, we have

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     Let ∠DBC=∠ACB=x         .......(1)AC=AD          (Given)It means that ∠ACD=∠BDC         (In triangle, angles opposite to equal sides are equal)     Let ∠ACD=∠BDC=y           ......(2)In ∆BDC, we have∠BDC+∠BCD+∠DBC=180°     (Angle sum property of triangle)

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     Let ∠DBC=∠ACB=x         .......(1)AC=AD          (Given)It means that ∠ACD=∠BDC         (In triangle, angles opposite to equal sides are equal)     Let ∠ACD=∠BDC=y           ......(2)In ∆BDC, we have∠BDC+∠BCD+∠DBC=180°     (Angle sum property of triangle)⇒∠BDC+∠ACB+∠ACD+∠DBC=180°

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     Let ∠DBC=∠ACB=x         .......(1)AC=AD          (Given)It means that ∠ACD=∠BDC         (In triangle, angles opposite to equal sides are equal)     Let ∠ACD=∠BDC=y           ......(2)In ∆BDC, we have∠BDC+∠BCD+∠DBC=180°     (Angle sum property of triangle)⇒∠BDC+∠ACB+∠ACD+∠DBC=180°Putting (1) and (2) in the above equation, we get

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     Let ∠DBC=∠ACB=x         .......(1)AC=AD          (Given)It means that ∠ACD=∠BDC         (In triangle, angles opposite to equal sides are equal)     Let ∠ACD=∠BDC=y           ......(2)In ∆BDC, we have∠BDC+∠BCD+∠DBC=180°     (Angle sum property of triangle)⇒∠BDC+∠ACB+∠ACD+∠DBC=180°Putting (1) and (2) in the above equation, we gety+x+y+x=180°

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     Let ∠DBC=∠ACB=x         .......(1)AC=AD          (Given)It means that ∠ACD=∠BDC         (In triangle, angles opposite to equal sides are equal)     Let ∠ACD=∠BDC=y           ......(2)In ∆BDC, we have∠BDC+∠BCD+∠DBC=180°     (Angle sum property of triangle)⇒∠BDC+∠ACB+∠ACD+∠DBC=180°Putting (1) and (2) in the above equation, we gety+x+y+x=180°⇒2x+2y=180°

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     Let ∠DBC=∠ACB=x         .......(1)AC=AD          (Given)It means that ∠ACD=∠BDC         (In triangle, angles opposite to equal sides are equal)     Let ∠ACD=∠BDC=y           ......(2)In ∆BDC, we have∠BDC+∠BCD+∠DBC=180°     (Angle sum property of triangle)⇒∠BDC+∠ACB+∠ACD+∠DBC=180°Putting (1) and (2) in the above equation, we gety+x+y+x=180°⇒2x+2y=180°⇒2(x+y)=180°

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     Let ∠DBC=∠ACB=x         .......(1)AC=AD          (Given)It means that ∠ACD=∠BDC         (In triangle, angles opposite to equal sides are equal)     Let ∠ACD=∠BDC=y           ......(2)In ∆BDC, we have∠BDC+∠BCD+∠DBC=180°     (Angle sum property of triangle)⇒∠BDC+∠ACB+∠ACD+∠DBC=180°Putting (1) and (2) in the above equation, we gety+x+y+x=180°⇒2x+2y=180°⇒2(x+y)=180°⇒(x+y)=180/2=90°

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     Let ∠DBC=∠ACB=x         .......(1)AC=AD          (Given)It means that ∠ACD=∠BDC         (In triangle, angles opposite to equal sides are equal)     Let ∠ACD=∠BDC=y           ......(2)In ∆BDC, we have∠BDC+∠BCD+∠DBC=180°     (Angle sum property of triangle)⇒∠BDC+∠ACB+∠ACD+∠DBC=180°Putting (1) and (2) in the above equation, we gety+x+y+x=180°⇒2x+2y=180°⇒2(x+y)=180°⇒(x+y)=180/2=90°Therefore, ∠BCD=90°

AB=AC         (Given)It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     Let ∠DBC=∠ACB=x         .......(1)AC=AD          (Given)It means that ∠ACD=∠BDC         (In triangle, angles opposite to equal sides are equal)     Let ∠ACD=∠BDC=y           ......(2)In ∆BDC, we have∠BDC+∠BCD+∠DBC=180°     (Angle sum property of triangle)⇒∠BDC+∠ACB+∠ACD+∠DBC=180°Putting (1) and (2) in the above equation, we gety+x+y+x=180°⇒2x+2y=180°⇒2(x+y)=180°⇒(x+y)=180/2=90°Therefore, ∠BCD=90°hope, this will help you.☺

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