Advertisements
Advertisements
Question
Prove that:
- ∆ ABD ≅ ∆ ACD
- ∠B = ∠C
- ∠ADB = ∠ADC
- ∠ADB = 90°
Advertisements
Solution
Given: In the figure,
AB = AC
BD = CD
To prove:
- Δ ABD ≅ Δ ACD
- ∠B = ∠C
- ∠ADB = ∠ADC
- ∠ADB = 90°
Proof: In Δ ABD and Δ ACD
AD = AD ...(common)
AB = AC ...(given)
BD = CD ...(given)
(i) ∴ Δ ABD ≅ Δ ACD ...(SSS axiom)
(ii) ∴ ∠B = ∠C ...(c.p.c.t.)
(iii) ∠ADB = ∠ADC ...(c.p.c.t.)
But ∠ADB + ∠ADC = 180° ...(Linear pair)
∴ ∠ADB = ∠ADC
(iv) ∠ADB = ∠ADC
= `(180°)/2`
= 90°
APPEARS IN
RELATED QUESTIONS
In a ΔABC, if AB = AC and ∠B = 70°, find ∠A.
In the pair of triangles in the following figure, parts bearing identical marks are congruent. State the test and the correspondence of vertices by the triangle in pairs is congruent.
In the pair of triangles in the following figure, parts bearing identical marks are congruent. State the test and the correspondence of vertices by the triangle in pairs is congruent.
In the pair of triangles given below, the parts shown by identical marks are congruent. State the test and the one-to-one correspondence of vertices by which the triangles in the pair are congruent, the remaining congruent parts.
The following figure shown a triangle ABC in which AB = AC. M is a point on AB and N is a point on AC such that BM = CN.
Prove that: ΔAMC≅ ΔANB
In the figure, AP and BQ are perpendiculars to the line segment AB and AP = BQ. Prove that O is the mid-point of the line segments AB and PQ.
ΔABC is isosceles with AB = AC. BD and CE are two medians of the triangle. Prove that BD = CE.
In ΔABC, AD is a median. The perpendiculars from B and C meet the line AD produced at X and Y. Prove that BX = CY.
Two right-angled triangles ABC and ADC have the same base AC. If BC = DC, prove that AC bisects ∠BCD.
∆ABC and ∆PQR are congruent under the correspondence:
ABC ↔ RQP
Write the parts of ∆ABC that correspond to
(i) `bar"PQ"`
(ii)∠Q
(iii) `bar"RP"`
