in triangle abc angle b = 2angle c.d is a point ob bc such that ad bisects angle bac and ab = cd.prove that angle bac = 72
There is a wonderful field of math lying ahead from this problem.
notice that cos 36° = .809016.. which is 1/2 of the golden ratio of (1+√5)/2 or 1.61803....
Look up the "pentagon"
Draw the central triangles, your triangle is one of these.
To prove that angle BAC is equal to 72 degrees, we can use the given information and apply the angle bisector theorem.
Here are the steps to prove it:
Step 1: Draw triangle ABC with angle BAC, angle ABC, and angle BCA.
Step 2: Label the point where line AD intersects side BC as point D.
Step 3: Since angle BAD bisects angle BAC, we have:
angle BAC = angle BAD + angle DAC (angle bisector theorem)
Step 4: Since AB = CD, we have:
AB/BC = CD/BC (sides opposite equal angles are proportional)
Step 5: Since angle B = 2 * angle C, we can set up the following equation:
AB/BC = BD/DC (angle bisector theorem)
Step 6: Substituting AB = CD, we have:
CD/BC = BD/DC
Step 7: Cross-multiplying, we get:
CD * DC = BD * BC
Step 8: Since AB = CD and AD bisects angle BAC, we can conclude that triangle BDA is congruent to triangle CDB by the side-angle-side (SAS) congruence criterion.
Step 9: Therefore, we can conclude that BD = DC.
Step 10: Substituting BD = DC into the equation from step 7, we get:
CD * CD = BD * BC
Step 11: Rearranging the equation, we have:
CD² = BD * BC
Step 12: Since BD = DC, we can rewrite the equation from step 11 as:
CD² = DC * BC
Step 13: Dividing both sides by DC, we get:
CD = BC
Step 14: Since CD = BC, we can conclude that angle BCA is an isosceles triangle, and therefore angle BAC is equal to angle BCA.
Step 15: Since angle BAC is equal to 2 * angle C, and angle C is one-third of angle BAC (since angle BAC = angle BAD + angle DAC), we can set up the following equation:
3 * angle C = angle BAC
Step 16: Substituting angle BAC = 2 * angle C, we get:
3 * angle C = 2 * angle C
Step 17: Dividing both sides by angle C, we have:
3 = 2
Step 18: This is a contradiction since 3 is not equal to 2.
Therefore, the statement that angle BAC = 72 degrees is false.
make a sketch.
let angle C = x , then angle B = 2x
also, let angle BAD = y = angle CAD , (Angle A was bisected by AD)
then in triangle ABC , 2x + 2y + x = 180
y = (180-3x)/2 = 90 - 3x/2
by exterior angle theorem,
angle BDA = x+y and angle CDA = 2x+y
let AB = m = DC , (given)
by the sine law:
m/sin(x+y) = AD/sin 2x and m/siny = AD/sinx
m = ADsin(x+y)/sin 2x and m = ADsiny/sinx
so ADsin(x+y)/sin2x = ADsiny/sinx
divide by AD
sin(x+y)/sin2x = siny/sinx
sin(x + 90-3x/2)/(2sinxcosx) = siny/sinx
sin(90 - x/2)/(2cosx) = siny , after dividing by sinx
cross-multiply
sin(90-x/2) = 2sinxcosx
sin(90-x/2) = sin 2x
so 90 - x/2 = 2x
times 2
180 - x = 4x
180=5x
x=36
then y = 90-3(36)/2 = 36
and angle A = 2y = 72°