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Areas of Parallelograms and Triangles
Question
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ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC. Show that ∠B = ∠C.

Answer

Given: ABC is an isosceles triangle with
AB = AC.
AP ⊥ BC
To Prove: ∠B = ∠C

Proof: In ∆ABC,
∵ AB = AC    | Given
∴ ∠ABC = ∠ACB    ...(1)
| Angles opposite to equal sides of a triangle are equal
Now, in ∆APB and ∆APC,
AB = AC    | Given
∠ABP = ∠ACP    | From (1)
∠APB = ∠APC (= 90°) | Given
∴ ∆APB ≅ ∆APC | AAS congruence rule
∴ ∠ABP = ∠ACP    | CPCT
⇒ ∠B = ∠C

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.

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.