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Triangles

Short Answer Type

481. In figure, AD π CD and BC π CD. If AQ = BP and DP = CQ, prove that ∠DAQ = ∠CBP.

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482. In ∆ABC, D is the mid point of BC. The perpendiculars from D to AB and AC are equal. Prove that ∆ABC is isosceles.

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483. In figure, ∆ABD and ∆BCD are isosceles triangles on the same base BD. Prove that ∠ABC = ∠ADC.

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Long Answer Type

484. In ∆’s ABC and PQR, AB = PQ, AC = PR and altitude AM and altitude PN are equal. Show that ∆ABC ≅ ∆PQR.

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485. In figure, the perpendicular AD, BE and CF drawn from the vertices A, B and C respectively of ∆ABC are equal. Prove that the triangle is an equilateral triangle.

Given: In figure, the perpendiculars AD, BE and CF drawn from the vertices A, B and C respectively of ∆ABC are equal.
To Prove: ∆ABC is an equilateral triangle.
Proof: In right triangles BEC and CFB,
Hyp. BC = Hyp. CB    | Common
Side BE = Side CF    | Given
∴ ∆BEC ≅ ∆CFB
| RHS congruence rule
∴ ∠BCE = ∠CBF    ...(1) | CPCT
In ∆ADB and ∆ADC,
∠ADB = ∠ADC (Each = 90°)
∠ABD = ∠ACD    | From (1)
AD = AD    | Common
∴ ∆ADB ≅ ∆ADC
| AAS congruence rule
∴ AB = AC    ...(2) | CPCT
Similarly, we can show that
AB = BC    ...(3)
and    AC = BC    ...(4)
From (2), (3) and (4), we get,
AB = BC = CA
⇒ ∆ABC is an equilateral triangle.

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486. ABCD is a parallelogram. If the two diagonals are equal, find the measure of ∠ABC.

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Short Answer Type

487. AB is a line-segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (see figure). Show that the line PQ is the perpendicular bisector of AB.

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488. P is a point equidistant from two lines l and m intersecting at point A (see figure). Show that the line AP bisects the angle between them.

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489.

In the figure, RS = QT and QS = RT. Prove that PQ = PR.

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490. In the given figure, D is the mid-point of base BC, DE and DF are perpendiculars to AB and AC respectively such that DE = DF. Prove that ∠B = ∠C.

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