In triangle ABC, AB=AC And BD = DC where D is a point on BC, then find ADB.
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Answer:
In △ABD,
∠ADB+∠A+∠ABD=180o
⇒∠ADB+36+36=180o
⇒∠ADB=180−72=108o
In an isosceles triangle , opposite to equal sides are equal. In a triangle, one side is extended,
the exterior angle is equal to sum of interior opposite angles
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see ur answer it's down ⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬⏬Given=> AB=AC
BD=DC
Construction => Join AD
To prove that => angle ABD = angle ACD
Proof=> In triangle ADB & ADC
AD=AD [Common]
BD=CD [Given]
AB=AC [Given] Triangle ADB is congruent to Triangle ADC.
Therefore,
Angle ABD = Angle ACD [ C.P.C.T ]
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