Isosceles Triangle Theorem
Answers
Step-by-step explanation:
two sides of a triangle are congruent , then the angles opposite to these sides are congruent.
∠P≅∠Q
Proof:
Let S be the midpoint of PQ¯¯¯¯¯ .
Join R and S .
Since S is the midpoint of PQ¯¯¯¯¯ , PS¯¯¯¯¯≅QS¯¯¯¯¯ .
By Reflexive Property ,
RS¯¯¯¯¯≅RS¯¯¯¯¯
It is given that PR¯¯¯¯¯≅RQ¯¯¯¯¯
Therefore, by SSS ,
ΔPRS≅ΔQRS
Since corresponding parts of congruent triangles are congruent,
∠P≅∠Q
The converse of the Isosceles Triangle Theorem is also true.
If two angles of a triangle are congruent, then the sides opposite those angles are congruent.
If ∠A≅∠B , then AC¯¯¯¯¯≅BC¯¯¯¯¯ .
Answer:
isosceles triangles are:
It has two equal sides.
It has two equal angles, that is, the base angles.
When the third angle is 90 degree, it is called a right isosceles triangle.
Theorem 1: Angles opposite to the equal sides of an isosceles triangle are also equal.
Proof: Consider an isosceles triangle ABC where AC = BC. We need to prove that the angles opposite to the sides AC and BC are equal, that is, ∠CAB = ∠CBA.
Isosceles Triangle
We first draw a bisector of ∠ACB and name it as CD.
Now in ∆ACD and ∆BCD we have,
AC = BC (Given)
∠ACD = ∠BCD (By construction)
CD = CD (Common to both)
Thus, ∆ACD ≅∆BCD (By SAS congruency)
So, ∠CAB = ∠CBA (By CPCTC)
Theorem 2: Sides opposite to the equal angles of a triangle are equal.
Proof: Consider an isosceles triangle ABC. We need to prove that AC = BC and ∆ABC is isosceles.
Isosceles Triangle Theorem 2
Now in ∆ACD and ∆BCD we have,
∠ACD = ∠BCD (By construction)
CD = CD (Common to both)
∠ADC = ∠BDC = 90° (By construction)
Thus, ∆ACD ≅ ∆BCD (By ASA congruency)
So, AB = AC (By CPCTC)
Or ∆ABC is isosceles.
Step-by-step explanation: