3 rd one how to solve it
Attachments:
Answers
Answered by
1
Hi friend,
Given: ABC is a triangle in which AB=AC. P is any point in the interior of the triangle such that angle ABP=ACP.
To Prove: AP bisects angle BAC
Proof:
In ∆APB and ∆APC,
AB = AC [given]
angleABP = angle ACP [given]
AP = AP [common]
∴tr. APB and tr. APC (by SAS)
also, angle PAB = angle PAC [corresponding angles of congruent triangles]
Given: ABC is a triangle in which AB=AC. P is any point in the interior of the triangle such that angle ABP=ACP.
To Prove: AP bisects angle BAC
Proof:
In ∆APB and ∆APC,
AB = AC [given]
angleABP = angle ACP [given]
AP = AP [common]
∴tr. APB and tr. APC (by SAS)
also, angle PAB = angle PAC [corresponding angles of congruent triangles]
Attachments:
Similar questions