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Standard VIII

Mathematics

SAS Postulate

Let A B C D b...

Question

Let ABCD be a parallelogram and suppose the bisectors of ∠A and ∠B meet at P. Prove that ∠APB = 90°.

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Solution

Given: Parallelogram ABCD, where the bisectors of angles A and B meet at P

To Prove:

Proof:

Since AP is the angle bisector of,.

Since BP is the angle bisector of,.

Since ABCD is a parallelogram, we get:

Now, in triangle APB, we have:

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Let ABCD be a parallelogram and suppose the bisectors of ∠A and ∠B meet at P. Prove that ∠APB = 90°.

Given: Parallelogram ABCD, where the bisectors of angles A and B meet at P To Prove: Proof: Since AP is the angle bisector of,. Since BP is the angle bisector of,. Since ABCD is a parallelogram, we get: Now, in triangle APB, we have:

Given: Parallelogram ABCD, where the bisectors of angles A and B meet at P

To Prove:

Proof:

Since AP is the angle bisector of,.

Since BP is the angle bisector of,.

Since ABCD is a parallelogram, we get:

Now, in triangle APB, we have: