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Areas of Parallelograms and Triangles
Question
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If the bisector of the vertical angle of a triangle bisects the base of the triangle, then prove that the triangle is isosceles.

Answer

Given: A ∆ABC in which the bisector of the vertical angle ∠BAC bisects the base BC, i.e., BD = CD

To Prove: ∆ABC is isosceles

Construction: Produce AD to E such that AD = DE. Join EC.
Proof: In ∆ADB and ∆EDC,
BD = CD    | Given
AD = ED    | By construction
∠ADB = ∠EDC
| Vertically opposite angles
∴ ∆ADB ≅ ∆EDC
| SAS congruence rule ∴ AB = EC    ...(1) | CPCT
and    ∠BAD = ∠CED    | CPCT
But ∠BAD = ∠CAD    | Given
∴ ∠CAD = ∠CED
∴ AC = CE    ...(2)
| Sides opposite to equal angles of a triangle are equal
From (1) and (2),
AB = AC
∴ ∆ABC is isosceles.

Sponsor Area

Some More Questions From Areas of Parallelograms and Triangles Chapter

In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

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.