Which of the following pairs of triangles are congruent by ASA condition?
1) We have
Since ∠ABO = ∠CDO = 45° and both are alternate angles,
∠BAO = ∠DCO (alternate angle , AB∥CD and AC is a transversal line)
∠ABO = ∠CDO = 45° (given in the figure)
Also, AB = DC (Given in the figure)
Therefore, by ASA △AOB ≅△DOC
In △ABC ,
Now AB = AC (Given)
∠ABD = ∠ACD = 40° (Angles opposite to equal sides)
∠ABD +∠ACD+∠BAC=180° (Angle sum property)
∠BAD =∠CAD = 50°
Therefore, by ASA, △ABD ≅△ADC
∠A+∠B+∠C=180°(Angle sum property)
∠P+∠Q+∠R=180°(Angle sum property)
∠BAC = ∠QPR = 30°
∠BCA=∠PRQ = 60°
and AC = PR (Given)
Therefore, by ASA, △ABC ≅△PQR
We have only BC =QR but none of the angles of △ABC AND △PQR are equal.
Therefore, △ABC≆ △PRQ
In Fig. 37, AD bisects ∠A and AD ⊥ BC.
(i) Is ∆ ADB ≅ ∆ ADC?
(ii) State the three pairs of matching parts you have used in (i).
(iii) Is it true to say that BD = DC?
(i) Yes, △ADB ≅△ADC, by ASA criterion of congruency
(ii) We have used ∠BAD =∠CAD
∠ADB=∠ADC = 90° since AD⊥BC
and AD = DA
(iii) Yes, BD= DC since, △ADB ≅△ADC
Draw any triangle ABC. Use ASA condition to construct another triangle congruent to it.
We have drawn
△ABC with ∠ABC = 60°and ∠ACB = 70°
We now construct △PQR≅△ABC
△PQR has ∠PQR =60° and ∠PRQ = 70°
Also we construct △PQR such that BC =QR
Therefore by ASA the two triangles are congruent
In ∆ ABC, it is known that ∠B = ∠C. Imagine you have another copy of ∆ ABC
(i) Is ∆ ABC ≅ ∆ ACB?
(ii) State the three pairs of matching parts you have used to answer (i).
(iii) Is it true to say that AB = AC?
(i) Yes△ABC ≅△ACB
(ii) We have used ∠ABC=∠ACB and ∠ACB =∠ABC again
Also BC = CB
(iii) Yes, it is true to say that AB = AC since ∠ABC=∠ACB
In Fig. 38, AX bisects ∠BAC as well as ∠BDC. State the three facts needed to ensure that ∆ ABD ≅ ∆ ACD.
As per the given conditions,∠CAD=∠BAD and ∠CDA=∠BDA (because AX bisects ∠BAC ) AD=DA (common)
Therefore, by ASA, △ACD≅△ABD
In Fig. 39, AO = OB and ∠A = ∠B.
(i) Is ∆ AOC ≅ ∆ BOD?
(ii) State the matching pair you have used, which is not given in the question.
(iii) Is it true to say that ∠ACO = ∠BDO?
∠OAC =∠OBD, AO = OBAlso, ∠AOC = ∠BOD (Opposite angles on same vertex)
Therefore, by ASA △AOC ≅△BOD