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Areas of Parallelograms and Triangles
Question
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In the given figure, if AB = FE, BC = ED, AB ⊥ BD and FE ⊥ EC, then prove that AD = FC.

Answer

Given:
AB = FE, BC = ED,
AB ⊥ BD and FE ⊥ EC
To Prove: AD = FC
Proof: In ∆ABD and ∆FEC,
AB = FE    ...(1) | Given
∠ABD = ∠FEC    ...(2)
| Each = 90°
BC = ED    | Given
⇒ BC + CD = ED + DC
⇒    BD = EC    ...(3)
In view of (1), (2) and (3),
∆ABD ≅ ∆FEC
| SAS congruence rule
∴ AD = FC    | CPCT

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Some More Questions From Areas of Parallelograms and Triangles Chapter

In quadrilateral ACBD, AC = AD and AB bisects ∠A (see figure). Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?

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.