Question

In a quadrilateral ABCD, if bisectors of the angle ABC and angle ADC meet on the diagonal AC, prove that the bisector of angle BAD and angle BCD will meet on the diagonal BD ​

Answer 1

According to question it is given that ABCD is a quadrilateral in which the bisectors of ∠

ABC and ∠

ADC meet on the diagonal AC at P.

TO PROVE: Bisectors of ∠

BAD and ∠

BCD meet on the diagonal BD

CONSTRUCTION: Join BP and DP. Let us suppose that  the bisector of ∠

BAD meet BD at Q. Now, Join AQ and CQ.

PROOF: In order to prove that the bisectors of ∠

BAD and ∠

BCD meet on the diagonal BD, we have to prove that CQ is a bisector of∠BCD for which we will prove that Q divides BD in the ratio BC: DC.

In Δ

ABC, BP is the bisector of ∠

ABC.( According to question)

∴ABBC=APPC

.......................(i)  

In Δ

ACD, DP is the bisector of ∠

ADC.(as per fig)

∴ADDC=APPC

.....................(ii)

Therefore, from (i) and (ii), we get

ABBC=ADDC

⇒ABAD=BCDC

.................... ...(iii)

Again, In Δ

ABD, AQ is the bisector of ∠

BAD. [By construction]

∴ABAD=BQDQ

.............(iv)

From (iii) and (iv), we get

BCDC=BQDQ.

Hence, in Δ

CBD, Q divides BD in the ratio of  CB: CD.

Thus, CQ is the bisector of ∠

BCD.

Therefore, the bisectors of ∠

BAD and ∠

BCD meet on the diagonal BD.( Hence proved)

Answer 2

Given : ABCD is  a quadrilateral and bisectors of ∠ABC and ∠ADC meet on the diagonal AC at P

To prove : bisectors of ∠BAD and ∠BCD meet on the diagonal BD

construction : join BP and DP

proof :

in triangle ABC , BP is the bisector of ∠ABC

therefore ,

$\frac{AB}{BC}= \frac{AP}{PC}...(1)$

in triangle ACD , DP is the bisector of ∠ADC

thus ,

$\frac{AD}{DC}= \frac{AP}{PC}...(2)$

from 1 and 2

$\frac{AB}{AD}= \frac{BC}{DC}...(3)$

Now , in triangle CBD , Q divides BD in teh ratio CB: CD

therefore ,

CQ is the bisector of ∠BCD

hence ,

Bisector of ∠BAD and ∠BCD meet on the diagonal BD

hence proved

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