Question

In the figure above ABCD is a parallelogram $\overline{DE},\overline{EF}$ and $\overline{BG}$ are the bisectors of $\angle ADC,\angle DEB$ and $\angle ABC$ respectively. BG and EF intersect at H. If $\angle DAB={70}^{\circ }$ then find $\angle BHE$

• A. ${50}^{\circ }$
• B. ${\left(62\frac{1}{2}\right)}^{\circ }$
• C. ${75}^{\circ }$
• D. ${\left(87\frac{1}{2}\right)}^{\circ }$

Answer 1

The correct option is B ${\left(62\frac{1}{2}\right)}^{\circ }$

Given $\angle BAD={70}^{\circ }$

So $\angle ADC=\angle ABC=180-70={110}^{\circ }$ (Sum of angles formed at adjacent vertices of parallelogram)

$\angle ADE = half of angle ADC =55 degree

For triangle ADE , exterior angle DEB = angle DAE + angle ADE

= 70 + 55 = ${125}^{\circ }$

Now angle HEB is half of angle DEB

Thus $\angle HEB=\frac{125}{2}={62.5}^{\circ }$

Angle HBC = half of angle ABC = ${55}^{\circ }$

In triangle HEB

$\angle EHB={180}^{\circ }-\angle HEB-\angle HBE$

=180 - 62.5- 55

= ${62.5}^{\circ }$

So the correct answer is option B