Advertisements
Advertisements
Question
In the figure, given below, triangle ABC is right-angled at B. ABPQ and ACRS are squares. 
Prove that:
(i) ΔACQ and ΔASB are congruent.
(ii) CQ = BS.
Advertisements
Solution
Given: A(Δ ABC) is right-angled at B.
ABPQ and ACRS are squares
To Prove:
(i) ΔACQ ≅ ΔASB
(ii) CQ = BS
Proof:
(i)
∠ QAB = 90° ...[ ABPQ is a square ] ...(1)
∠ CAS = 90° ...[ ACRS is a square ] ...(2)
From (1) and (2) , We have
∠ QAB = ∠CAS ...(3)
Adding ∠BAC to both sides of (3), We have
∠ QAB + ∠BAC = ∠CAS+ ∠BAC
⇒ ∠QAC = ∠BAS ...(4)
In ΔACQ ≅ ΔASB, (by SAS)
QA = AB ...[ Sides of a square ABPQ ]
∠QAC = ∠SAB ...[ From(4) ]
AC = AS ...[ sides of a square ACRS ]
∴ By Side -Angle-Side criterion of congruence,
ΔACQ ≅ ΔASB
(ii)
The corresponding parts of the congruent triangles are congruent,
∴ CQ = SB ...[ c.p.c.t. ]
APPEARS IN
RELATED QUESTIONS
In quadrilateral ACBD, AC = AD and AB bisects ∠A (See the given figure). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?

ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (See the given figure). Prove that
- ΔABD ≅ ΔBAC
- BD = AC
- ∠ABD = ∠BAC.

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

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 the given figure). Show that:
- ΔAMC ≅ ΔBMD
- ∠DBC is a right angle.
- ΔDBC ≅ ΔACB
- CM = `1/2` AB

In the figure, the two triangles are congruent.
The corresponding parts are marked. We can write ΔRAT ≅ ?

Which of the following statements are true (T) and which are false (F):
If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.
If ABC and DEF are two triangles such that AC = 2.5 cm, BC = 5 cm, ∠C = 75°, DE = 2.5 cm, DF = 5cm and ∠D = 75°. Are two triangles congruent?
If the following pair of the triangle is congruent? state the condition of congruency:
In ΔABC and ΔQRP, AB = QR, ∠B = ∠R and ∠C = P.
A triangle ABC has ∠B = ∠C.
Prove that: The perpendiculars from the mid-point of BC to AB and AC are equal.
ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
