Question

ABCDE is a regular pentagon. The bisector
of D meets its opposite side AB at point P.
Find the measure of DPB.​

Answer 1

Each interior angle of a regular pentagon

$= \frac{2n - 4}{n} \times 90$

$= \frac{2 \times 5 - 4}{5} \times {90}^{o}$

$=6×18= {108}^{o}$

$∴∠BDP= \frac{1}{2} ( {108}^{o}) = {54}^{o}$

In the quadrilateral ABCM,

∠BAM+∠ABC+∠BCM+∠AMC=360°

⇒54°+108°+108°+∠AMC=360°

⇒DPB=90°