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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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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.

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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
$left parenthesis iv right parenthesis space space space space CM equals 1 half increment ACB$

Given: 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.

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

(i) In ∆AMC and ∆BMD,
AM = BM
| ∵ M is the mid-point of the hypotenuse AB
CM = DM    | Given
∠AMC = ∠BMD
| Vertically Opposite Angles
∴ ∆AMC ≅ ∆BMD. | SAS Rule
(ii)    ∵ ∆AMC ≅ ∆BMD
| From (i) above
∠ACM = ∠BDM    | C.P.C.T.
But these are alternate interior angles and they are equal
∴ AC || BD
Now, AC || BD and a transversal BC intersects them
∴ ∠DBC + ∠ACB = 180°
| ∵ The sum of the consecutive interior angles on the same side of a transversal is
180°
⇒ ∠DBC + 90° = 180°
| ∵ ∠ACB = 90° (given)
⇒     ∠DBC = 180° - 90° = 90°
⇒ ∠DBC is a right angle.
(iii)    In ∆DBC and ∆ACB,
∠DBC = ∠ACB (each = 90°)
| Proved in (ii) above
BC = CB    | Common
∵ ∆AMC ≅ ∆BMD    | Proved in (i) above
∴ AC = BD    | C.P.C.T.
∴ ∆DBC ≅ ∆ACB.    | SAS Rule
(iv)    ∵ ∆DBC ≅ ∆ACB
| Proved in (iii) above
∴ DC = AB    | C.P.C.T.

$rightwards double arrow$ 2CM = AB

   $vertical line space because space space DM equals CM equals 1 half DC$
$rightwards double arrow space space space space space CM equals 1 half AB.$