Question

26. In the given figure, AP and CP are bisectors of A and C respectively and / || m. Find the measure
of APC​

Answer 1

Answer:

Step-by-step explanation

Angle lAC + Angle mCA = 180* (Interior angles on the same side of the transversal)

1/2 angle lAC + 1/2 Angle mCA =90*

Angle PAC + Angle PCA = 90*

There fore Angle APC = 180* - 90* = 90* (Sum of the angles of a triangle is 180*)

Angle APC is 90*

Answer 2

Answer:

Measure of APC​ is $90^{0}$

Step-by-step explanation:

Given:

Here AP and CP are the bisectors of $A and C$ respectively and also $L$parallel to $\mathrm{M}$.

$\angle \mathrm{ACC}+\angle \mathrm{ACM}=180^{\circ} \quad [co interior angles]$

Dividing both the sides by 2 , we get

$\frac{\angle L A C}{2}+\frac{\angle A C M}{2}\\=\frac{180}{2} \\ \angle L A C+\angle A C P=90^{\circ}$  [Given]===>[1]

Now, in $\triangle ACP$

$\angle P A C+\angle A C P+\angle A P C=180^{\circ} \quad [Angle sum property]\\90^{\circ}+\angle A P C=180^{\circ}\\ \angle A P C=180^{\circ}-90^{\circ} \\ \angle A P C=90^{\circ}$