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Areas of Parallelograms and Triangles
Question
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ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively (see figure). Show that these altitudes are equal.

Answer

Given: ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively.
To Prove: BE = CF.
Proof: ∵ ABC is an isosceles triangle
∴ AB = AC
∴ ∠ABC = ∠ACB    ...(1)
| Angles opposite to equal sides of a triangle are equal
In ∆BEC and ∆CFB,
∠BEC = ∠CFB    | Each = 90°
BC = CB    | Common
∠ECB = ∠FBC    | From (1)
∴ ∆BEC ≅ ∆CFB    | By AAS Rule
∴ BE = CF.    | C.P.C.T.

Sponsor Area

Some More Questions From Areas of Parallelograms and Triangles Chapter

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.

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.