Question

ABCD is a parallelogram with ∠A = 80°. The internal bisectors of ∠B and ∠C

meet each other at O. Find the measure of the three angles of ΔBCO.​

Answer 1

Step-by-step explanation:

Since, it is a parallelogram,

∠A =∠C = 80° (Opposite angles of a parallelogram are equal)

∠A + ∠B = 180° (Adjacent angles of a parallelogram are supplementary)

$= > ∠B = 180 - 80 = 100$

Now

$∠D = ∠B = {100}^{o}$

Now in ΔBCO

$bisector \: of \: ∠B + bisector \: of \: ∠C + ∠O = 180 \: degrees \\ = > 50 + 40 + ∠O = 180 \\ = > ∠O = 90 \: degrees$

∠O = 90°

∠OBC = 50°

∠OCB = 40°

Om Namah Shivaya

Debasish Biswal