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Areas of Parallelograms and Triangles
Question
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In figure, AB = AC, D is the point in the interior of ∆ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC of ∆ABC.

Answer

Given: AB = AC, D is the point in the interior of ∆ABC such that ∠DBC = ∠DCB.
To Prove: AD bisects ∠BAC of ∆ABC.
Proof: In ∆DBC,
∵ ∠DBC = ∠DCB ...(1) | Given
∴ DB = DC    ...(2)
| Sides opposite to equal angles of a triangle are equal
In ∆ABC,
∵ AB = AC    | Given
∴ ∠ABC = ∠ACB
∴ ∠ABC = ∠ACB    ...(3)
| Angles opposite to equal sides of a triangle are equal
Subtracting (1) from (3), we get,
∠ABC - ∠DBC = ∠ACB - ∠DCB
⇒ ∠ABD = ∠ACD    ...(4)
In ∆ADB and ∆ADC,
AB = AC    | Given
DB = DC    | Proved in (2)
∠ABD = ∠ACD | Proved in (4)
∴ ∆ADB ≅ ∆ADC
| SAS congruence rule
∴ ∠DAB = ∠DAC    | CPCT
⇒ AD bisects ∠BAC of ∆ABC.

Sponsor Area

Some More Questions From Areas of Parallelograms and Triangles Chapter

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see figure). Show that.

(i) ∆DAP ≅ ∆EBP
(ii) AD = BE.

 

In figure, PS = QR and ∠SPQ = ∠RQP. Prove that PR = QS and ∠QPR = ∠PQS.

In figure, AP and BQ are perpendiculars to the line-segment AB and AP = BQ. Prove that O is the midpoint of line segments AB and PQ.

In figure, diagonal AC of a quadrilateral ABCD bisects the angles A and C. Prove that AB = AD and CB = CD.

AB is a line-segment. AX and BY are two equal line-segments drawn on opposite sides of line AB such that AX || BY. If AB and XY intersect each other at P. Prove that:
(i)    ∆APX ≅ ∆BPY
(ii)    AB and XY bisect each other at P.

In figure, ∠QPR = ∠PQR and M and N are respectively points on sides QR and PR of ∆PQR, such that QM = PN. Prove that OP = OQ, where O is the point of intersecting of PM and QN.

In figure, ∠B = ∠.E, BD = CE and ∠1 = ∠2. Show ∆ABC ≅ ∆AED.

In figure given below, AD is the median of ∆ABC.
BE ⊥ AD, CF ⊥ AD. Prove that BE = CF.