Advertisements
Advertisements
Question
ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC.
Advertisements
Solution
Given: In right angled ΔABC, AB = AC and CD is the bisector of ∠C.
Construction: Draw DE ⊥ BC.
To prove: AC + AD = BC
Proof: In right angled ΔABC, AB = AC and BC is a hypotensue ...[Given]
∴ ∠A = 90°
In ΔDAC and ΔDEC, ∠A = ∠3 = 90°
∠1 = ∠2 ...[Given, CD is the bisector of ∠C]
DC = DC ...[Common sides]
∴ ΔDAC ≅ ΔDEC ...[By AAS congruence rule]
⇒ DA = DE [By CPCT] ...(i)
And AC = EC ...(ii)
In ΔABC, AB = AC
∠C = ∠B [Angles opposite to equal sides are equal] ...(iii)
Again, in ΔABC, ∠A + ∠B + ∠C = 180° ...[By angles sum property of a triangle]
⇒ 90° + ∠B + ∠B = 180° ...[From equation (iii)]
⇒ 2∠B = 180° – 90°
⇒ 2∠B = 90°
⇒ ∠B = 45°
In ΔBED, ∠5 = 180° – (∠B + ∠4) ...[By angle sum property of a triangle]
= 180° – (45° + 90°)
= 180° – 135°
= 45°
∴ ∠B = ∠5
⇒ DE = BE [∵ Sides opposite to equal angles are equal] ...(iv)
From equations (i) and (iv),
DA = DE = BE ...(v)
∵ BC = CE + EB
= CA + DA ...[From equations (ii) and (v)]
∴ AD + AC = BC
Hence proved.
APPEARS IN
RELATED QUESTIONS
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see the given figure). Show that:
- ΔABM ≅ ΔPQN
- ΔABC ≅ ΔPQR
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
ABC is an isosceles triangle with AB = AC. Drawn AP ⊥ BC to show that ∠B = ∠C.
In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
In the following figure, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that ∆ABC ≅ ∆DEF.
ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC and AC. Prove that AE = BD.
Two lines l and m intersect at the point O and P is a point on a line n passing through the point O such that P is equidistant from l and m. Prove that n is the bisector of the angle formed by l and m.
Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.
ABCD is a quadrilateral such that diagonal AC bisects the angles A and C. Prove that AB = AD and CB = CD.
ABCD is quadrilateral such that AB = AD and CB = CD. Prove that AC is the perpendicular bisector of BD.
