Areas of Parallelograms and Triangles

Question

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.

Answer

Given: 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.
To Prove: (i) ∆DAP ≅ ∆EBP
(ii) AD = BE.
Proof: (i) In ∆DAP and ∆EBP,
AP = BP
| ∵ P is the mid-point of the line segment AB
∠DAP = ∠EBP    | Given
∠EPA = ∠DPB    | Given
⇒ ∠EPA + ∠EPD = ∠EPD + ∠DPB
| Adding ∠EPD to both sides
⇒    ∠APD = ∠BPE
∴ ∠DAP ≅ ∠EBP    | ASA Rule
(ii) ∵ ∆DAP ≅ AEBP    | From (i) above
∴ AD = BE.    | C.P.C.T.

Sponsor Area

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.