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Areas Of Parallelograms And Triangles

Question
CBSEENMA9002601
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In an isosceles triangle ABC with AB = AC, BD and CE are two medians. Prove that BD = CE. 

Solution

Given: In an isosceles triangle ABC with AB = AC, BD and CE are two medians.
To Prove: BD = CE

Proof: In ∆ABC,
∵ AB = AC
∴ ∠BC = ∠ACB    ...(1)
| Angles opposite to equal sides of a triangle are equal
Also,  1 half AB equals 1 half AC

| Halves of equals are equal ⇒ BE = CD    ...(2)

| ∵ BD and CE are two medians
Now, in ∆BDC and ∆CEB,
∠BCD = ∠CBE    | From (1)
BE = CD    | From (2)
BC = CB    | Common
∴ ∆BDC ≅ ∆CEB
| SAS congruence rule
∴ BD = CE    | CPCT

 

Some More Questions From Areas of Parallelograms and Triangles Chapter

ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see figure). Prove that:

(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.

AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

I and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ABC ≅ ∆CDA.

Line I is the bisector of an angle ∠A and B is any point on I. BP and BQ are perpendiculars from B to the arms of ∠A (see figure). Show that:
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.

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.

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