Q29 - Incenter, cyclic quadrilateral
Q29 - In quadrilateral ABCD, ∠DAB = ∠ABC = 110º, ∠BCD = 35º, ∠CDA = 105º and AC̅ bisects ∠DAB. Find ∠ABD.
Answer: 40 degree.
Solution:
1. Build an auxiliary point
Pick a point E on BC so that $$ DE\;||\;AB . $$ Because \(\angle BAD=\angle ABE=110^{\circ}\) and the two bases are parallel, ABED is an isosceles trapezoid, hence cyclic. Consequently
\(\angle ABD=\angle AED\) because same chord AD in a cyclic quadrilateral.
Triangle \(CDE\) is isosceles.
Parallelism gives \(\angle DEC=\angle ABE=110^{\circ}.\) Using the given \(\angle BCD=35^{\circ}\), $$ \angle CDE =180^{\circ}-\angle DEC-\angle ECD =180^{\circ}-110^{\circ}-35^{\circ} =35^{\circ}. $$ Thus \(\angle CDE=\angle ECD\), and $$ \boxed{\;CE=DE\;} $$ so C and D sit on a circle with centre E and radius CE=DE. 3. Put A on that same circle Because AC bisects \(\angle DAB\), $$ \angle CAD=\tfrac12\angle DAB =\tfrac12\,(110^{\circ}) =55^{\circ}. $$ In the circle just found, the central angle \(\angle CED\) equals \(110^{\circ}\). An inscribed angle that subtends the same chord CD must therefore be half of that, i.e. \(55^{\circ}\). Since \(\angle CAD\) is exactly \(55^{\circ}\), point A lies on the circle as well. So A,C,D are concyclic with centre E.
4. Transfer the 20° into the desired angle
Inside that same circle, $$ \angle ACD = 180^{\circ}-\angle CAD-\angle CDA = 180^{\circ}-55^{\circ}-105^{\circ} = 20^{\circ}. $$ For a given chord AD,
$$ \text{central angle } \angle AED = 2 \times \text{inscribed angle } \angle ACD = 2 \times 20^{\circ} = 40^{\circ}. $$
5. Finish
From Step 1, \(\angle ABD=\angle AED\).
From Step 4, \(\angle AED = 40^{\circ}\).
$$ \boxed{\angle ABD = 40^{\circ}} $$
Answer: 40 degree.
Solution:
Quick Solution:
Let I denote the incenter of △ABD. Then quadrilateral IBCD is cyclic since
∠DIB = 90º + ½ ∠DAB = 145º. Hence we obtain
∠IBD = ∠ICD = 180º − (55º + 105º) = 20º and ∠ABD = 40º.
Detailed solution:
Step 1:
If I is the incenter then ∠DIB = 90º + ½ ∠DAB = 145º. Why?
Step 1:
If I is the incenter then ∠DIB = 90º + ½ ∠DAB = 145º. Why?
Take any triangle XYZ. Let's say the angle bisectors of X and Z meet at O.
Then prove that ∠XOZ = 90º + ½ ∠XYZ. It's quite simple. Since I is the incenter here it means all the angle bisectors of A,B,D in △ABD meet at I. So this result will be easy to prove.
Step 2:
IBCD is a cyclic quadrilateral. Why?
We just need to prove that 1 pair of opposite angles sums up to 180º. Since ∠I = 145º and ∠I = 35º, it's proved.
We just need to prove that 1 pair of opposite angles sums up to 180º. Since ∠I = 145º and ∠I = 35º, it's proved.
Step 3:
∠IBD = ∠ICD why? Since in a circle angles subtended by the same chord are equal and both these angles are subtended by the chord ID.
Now ∠ICD = 180º − (55º + 105º) = 20º because ∠ICD is same as ∠ACD and we know all the other angles in △ACD.
Now ∠ICD = 180º − (55º + 105º) = 20º because ∠ICD is same as ∠ACD and we know all the other angles in △ACD.
Step 4:
Since ∠IBD = 20º and it's bisecting ∠ABD, ∠ABD = 40º.
Another Solution
1. Build an auxiliary point
Pick a point E on BC so that $$ DE\;||\;AB . $$ Because \(\angle BAD=\angle ABE=110^{\circ}\) and the two bases are parallel, ABED is an isosceles trapezoid, hence cyclic. Consequently
\(\angle ABD=\angle AED\) because same chord AD in a cyclic quadrilateral.
Triangle \(CDE\) is isosceles.
Parallelism gives \(\angle DEC=\angle ABE=110^{\circ}.\) Using the given \(\angle BCD=35^{\circ}\), $$ \angle CDE =180^{\circ}-\angle DEC-\angle ECD =180^{\circ}-110^{\circ}-35^{\circ} =35^{\circ}. $$ Thus \(\angle CDE=\angle ECD\), and $$ \boxed{\;CE=DE\;} $$ so C and D sit on a circle with centre E and radius CE=DE. 3. Put A on that same circle Because AC bisects \(\angle DAB\), $$ \angle CAD=\tfrac12\angle DAB =\tfrac12\,(110^{\circ}) =55^{\circ}. $$ In the circle just found, the central angle \(\angle CED\) equals \(110^{\circ}\). An inscribed angle that subtends the same chord CD must therefore be half of that, i.e. \(55^{\circ}\). Since \(\angle CAD\) is exactly \(55^{\circ}\), point A lies on the circle as well. So A,C,D are concyclic with centre E.
4. Transfer the 20° into the desired angle
Inside that same circle, $$ \angle ACD = 180^{\circ}-\angle CAD-\angle CDA = 180^{\circ}-55^{\circ}-105^{\circ} = 20^{\circ}. $$ For a given chord AD,
$$ \text{central angle } \angle AED = 2 \times \text{inscribed angle } \angle ACD = 2 \times 20^{\circ} = 40^{\circ}. $$
5. Finish
From Step 1, \(\angle ABD=\angle AED\).
From Step 4, \(\angle AED = 40^{\circ}\).
$$ \boxed{\angle ABD = 40^{\circ}} $$
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