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Triangles

Short Answer Type

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

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2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see figure). Prove that:

(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.

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3. AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

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4. I and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ABC ≅ ∆CDA.

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5.
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.

Given: Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A.
To Prove: (i) ∆APB ≅ ∆AQB
(ii) BP = BQ
Or
B is equidistant from the arms of ∠A.
Proof: (i) In ∆APB and ∆AQB,
∠BAP = ∠BAQ
| ∵ l is the bisector of ∠A
AB = AB    | Common
∠BPA = ∠BQA    | Each = 90°
| ∵ BP and BQ are perpendiculars from B to the arms of ∠A
∴ ∆APB ≅ ∆AQB    | AAS Rule
(ii) ∵ ∆APB ≅ ∆AQB
| Proved in (i) above
∴ BP = BQ.    | C.P.C.T.

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6. In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

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7. 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.

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Long Answer Type

8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that:

(i)    ∆AMC ≅ ∆BMD
(ii)    ∠DBC is a right angle
(iii)    ∆DBC ≅ ∆ACB
$left parenthesis iv right parenthesis space space space space CM equals 1 half increment ACB$