$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ (see Fig). Show that $\Delta ABC \cong \Delta CDA$.
(N/A) Given: $l \parallel m$ and $p \parallel q$.
To prove: $\Delta ABC \cong \Delta CDA$.
Proof:
Since $l \parallel m$ and $AC$ is a transversal,
$\therefore \angle BAC = \angle DCA$ [Alternate interior angles]
Also,since $p \parallel q$ and $AC$ is a transversal,
$\therefore \angle BCA = \angle DAC$ [Alternate interior angles]
Now,in $\Delta ABC$ and $\Delta CDA$:
$\angle BAC = \angle DCA$ [Proved above]
$\angle BCA = \angle DAC$ [Proved above]
$AC = CA$ [Common side]
Therefore,by the $ASA$ (Angle-Side-Angle) congruence criterion,$\Delta ABC \cong \Delta CDA$.
To prove: $\Delta ABC \cong \Delta CDA$.
Proof:
Since $l \parallel m$ and $AC$ is a transversal,
$\therefore \angle BAC = \angle DCA$ [Alternate interior angles]
Also,since $p \parallel q$ and $AC$ is a transversal,
$\therefore \angle BCA = \angle DAC$ [Alternate interior angles]
Now,in $\Delta ABC$ and $\Delta CDA$:
$\angle BAC = \angle DCA$ [Proved above]
$\angle BCA = \angle DAC$ [Proved above]
$AC = CA$ [Common side]
Therefore,by the $ASA$ (Angle-Side-Angle) congruence criterion,$\Delta ABC \cong \Delta CDA$.
Explore More
Similar Questions
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 Fig). Show that:
$(i)$ $\Delta AMC \cong \Delta BMD$
$(ii)$ $\angle DBC$ is a right angle.
$(iii)$ $\Delta DBC \cong \Delta ACB$
$(iv)$ $CM = \frac{1}{2} AB$
$(i)$ $\Delta AMC \cong \Delta BMD$
$(ii)$ $\angle DBC$ is a right angle.
$(iii)$ $\Delta DBC \cong \Delta ACB$
$(iv)$ $CM = \frac{1}{2} AB$
Difficult
View SolutionIn quadrilateral $ACBD$,$AC = AD$ and $AB$ bisects $\angle A$ (see figure). Show that $\Delta ABC \cong \Delta ABD$. What can you say about $BC$ and $BD$?
Medium
View SolutionShow that the angles of an equilateral triangle are $60^o$ each.
Difficult
View Solution$BE$ and $CF$ are two equal altitudes of a triangle $ABC$. Using $RHS$ congruence rule,prove that the triangle $ABC$ is isosceles.
Difficult
View Solution$\Delta ABC$ and $\Delta DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig). If $AD$ is extended to intersect $BC$ at $P$,show that:
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo