in a Quadrilateral ABCD, AB=AD and CB=CD.
Prove that-
(1) AC bisects angle BAD
(2) AC is perpendicular bisector of BD
Draw diagonal BD
Then triangles ABD and CBD are isosceles.
Hence, their altitudes bisect their bases and their vertex angles.
Plz tell me answer
To prove both (1) and (2), we can use the properties of a parallelogram. Since the given quadrilateral ABCD has AB = AD and CB = CD, it can be concluded that ABCD is a parallelogram.
Now let's prove (1):
We want to show that AC bisects angle BAD.
To do that, we need to prove that angle BAC is equal to angle DAC.
To start, recall that in a parallelogram, opposite angles are equal.
In the parallelogram ABCD, opposite angles are angle ABC and angle ADC, and angle BAD and angle BCD.
Now, since ABCD is a parallelogram, we have:
angle ABC = angle ADC (opposite angles in a parallelogram)
angle BCD = angle BAC (opposite angles in a parallelogram)
From the given information, AB = AD, and by the reflexive property of equality, AD = AB.
So, triangle ABD is an isosceles triangle, which means angles BDA and BAD are equal.
Therefore, we have:
angle BDA = angle BAD
Substituting this value into the equation above:
angle BCD = angle BAD (since angle BDA = angle BAD)
Now, combining the equations:
angle BCD = angle BAC (opposite angles in a parallelogram)
angle BAC = angle BAD (from earlier substitution)
Hence, we have proven that angle BAC is equal to angle DAC, which means AC bisects angle BAD.
Now let's prove (2):
We want to show that AC is the perpendicular bisector of BD.
To do that, we need to prove that AC is perpendicular to BD and AC bisects BD.
To start, recall that in a parallelogram, opposite sides are equal in length.
In the parallelogram ABCD, opposite sides are AB and CD, and BC and AD.
From the given information, AB = AD, and CB = CD.
Now, since AB = AD, and CB = CD, and AC is the diagonal of the parallelogram, we have:
Triangle ABD and triangle BCD are congruent by Side-Side-Side (SSS) congruence.
And, in congruent triangles, corresponding angles are equal.
Therefore, angle BAD = angle BCD.
We already proved that AC bisects angle BAD, so angle DAC = angle BAD/2.
Since angle BAD = angle BCD, we have:
angle DAC = angle BCD/2
Now, let's consider triangle BCD.
From SSS congruence, we know that triangle BCD is congruent to triangle BCD (by the reflexive property).
Therefore, angle BCD = angle BDC.
Since angle DAC = angle BCD/2, we have:
angle BDC = angle DAC.
Hence, AC is perpendicular to BD and AC bisects BD.
Therefore, we have proven both (1) and (2) - AC bisects angle BAD and AC is the perpendicular bisector of BD in the given quadrilateral ABCD.