Question

ABCD is a parallelogram DE, EF, BG are the bisectors if angle ADC, angle DEB and ABC respectively .BG and EF intersect at H. If angle DAB =70° then find angle BHE.​

Answer 1

Answer:

 the below figure bisectors of ∠B AND ∠D of a quadrilateral ABCD meets CD and AB , produced to P and Q respectively . prove that ∠P + ∠Q = (∠ABC + ∠ADC) . please solve this

Answer 2

Given:

ABCD is a parallelogram.

$\red{DE\ is\ the\ bisector\ of \angle ADC}$

$\blue{EF\ is\ the\ bisector\ of \angle DEB}$

$\green{BG\ is\ the\ bisector\ of \angle ABC}$

BG and EF intersect at H

Also, ∠DAB =70°

[See the attached image of the parallelogram drawn according to the question.]

Now,

Here are some properties of a parallelogram that will help to find the required answer-

• The opposite sides of a parallelogram are equal.
• The opposite angles are equal.
• The sum of two adjacent angles is 180°.
• The opposite sides are parallel.

Now,

Given that ∠DAB =70°

So,

∠ADC + ∠DAB = 180°

[ ∴ Sum of the two adjacent angles of a parallelogram is 180°]

⇒ ∠ADC + 70° = 180°

⇒ ∠ADC = 180° - 70°

⇒ ∠ADC = 110°

As

$\red{DE\ is\ the\ bisector\ of \angle ADC}$

So, ∠ADE = ∠EDC

∠ADE + ∠EDC = 110°

→ ∠ADE + ∠ADE = 110°

→ 2∠ADE = 110°

→ ∠ADE = 110° ÷ 2

→ ∠ADE = 110 ÷ 2 = 55°

So,

∠ADE = ∠EDC = 55° each

Now,

We can find a triangle ADE,

Exterior ∠DEB = ∠DAE + ∠ADE (∵ exterior angle property of a triangle)

⇒ ∠DEB = 70° + 55°

⇒ ∠DEB = 125°

Now,

$\blue{EF\ is\ the\ bisector\ of \angle DEB}$ (given)

So, ∠BEH = ∠DEH

→ ∠BEH + ∠DEH = 125°

→ ∠BEH + ∠BEH = 125°

→ 2∠BEH= 125°

→ ∠BEH = 125° ÷ 2

→ ∠BEH = 125 ÷ 2 = 62.5°

So,

∠BEH = ∠DEH = 62.5° each

Now,

We can also get that,

∠ADC = ∠ABC (Opposite angles of the Parallelogram ABCD)

∠ABC = 110°

Also,

$\green{BG\ is\ the\ bisector\ of \angle ABC}$ (given)

So,

∠HBE = ∠HBC = 55° each  (110° ÷ 2)

In triangle HBE

We have

∠HBE = 55° and ∠BEH = 62.5°

We know that sum of all three interior angles of triangle = 180°

[∵ Angle sum property.]

So,

∠HBE + ∠BEH + ∠BHE = 180°

⇒ 55° + 62.5° + ∠BHE = 180°

⇒ 117.5° + ∠BHE = 180°

⇒ ∠BHE = 180° - 117.5°

⇒ ∠BHE = 62.5°

Hence,

The value of ∠BHE = 62.5°