AP bisects angle A-

Question

AP bisects angle A-

Toppr Admin · Accepted Answer

Given that ABCD is a parallelogram-xA0-The bisector of -x2220-A- bisect CD at PAB-CD -opposite sides of a parallelogram- and AE is transversal-x21D2-x2220-BAP-x2220-DPA -alternate angles-As AP bisect -x2220-A-xA0-x21D2-x2220-DAP-x2220-BAP

$A P$ bisects angle $A$ .

Given that $A B C D$ is a parallelogram
The bisector of $∠ A$ , bisect $C D$ at $P$
$A B | | C D$ [opposite sides of a parallelogram] and $A E$ is transversal
$\Rightarrow ∠ B A P = ∠ D P A$ [alternate angles]
As $A P$ bisect $∠ A$ ,
$\Rightarrow ∠ D A P = ∠ B A P$

AP bisects angle A.

Given that ABCD is a parallelogram The bisector of ∠A, bisect CD at PAB||CD [opposite sides of a parallelogram] and AE is transversal⇒∠BAP=∠DPA [alternate angles]As AP bisect ∠A, ⇒∠DAP=∠BAP

$A P$ bisects angle $A$ .

Given that $A B C D$ is a parallelogram
The bisector of $∠ A$ , bisect $C D$ at $P$
$A B | | C D$ [opposite sides of a parallelogram] and $A E$ is transversal
$\Rightarrow ∠ B A P = ∠ D P A$ [alternate angles]
As $A P$ bisect $∠ A$ ,
$\Rightarrow ∠ D A P = ∠ B A P$