find the missing lengths and angle measures in kite abcd.
1. ac
2. cd
3. ( mangle abe )
4. ( mangle bce )

Question

Accepted Answer

# Explanation:
## Step1: Find AC
In a kite, the diagonals are perpendicular, and one diagonal is bisected. Here, $ AE = EC $ (since diagonals of a kite bisect each other? Wait, no, in a kite, one diagonal is bisected. Wait, actually, in a kite, one diagonal is the perpendicular bisector of the other. Wait, looking at the diagram, $ E $ is the intersection of diagonals. In a kite, one diagonal (the axis of symmetry) bisects the other. So if $ EC = 20 $, then $ AE = 20 $, so $ AC = AE + EC = 20 + 20 = 40 $.
## Step2: Find CD
In a kite, adjacent sides are equal. So $ CD = BC $? Wait, no, in a kite, two distinct pairs of adjacent sides are equal. So $ AB = AD $ and $ BC = CD $? Wait, no, actually, in a kite, $ AB = BC $ and $ AD = CD $? Wait, no, the definition is two pairs of adjacent sides are equal. So $ AB = AD $ and $ BC = CD $? Wait, looking at the diagram, $ DE = 21 $, $ EC = 20 $. Wait, $ CD $ is a side, and triangle $ CDE $ is a right triangle (since diagonals in a kite are perpendicular). So $ CD = \sqrt{DE^2 + EC^2} $? Wait, no, wait, $ DE = 21 $, $ EC = 20 $, so $ CD = \sqrt{21^2 + 20^2} $? Wait, no, wait, in the kite, $ AD = CD $? Wait, no, $ AB = BC $ and $ AD = CD $? Wait, actually, in a kite, one diagonal is the perpendicular bisector, so triangles $ ABE $ and $ ADE $ are congruent, and triangles $ BCE $ and $ DCE $ are congruent. So $ CD = BC $? Wait, no, $ CD $ is equal to $ AD $? Wait, no, let's check the right triangles. $ DE = 21 $, $ EC = 20 $, so $ CD = \sqrt{DE^2 + EC^2} $? Wait, no, $ DE $ is 21, $ EC $ is 20, so $ CD = \sqrt{21^2 + 20^2} = \sqrt{441 + 400} = \sqrt{841} = 29 $. Wait, because $ 21^2 = 441 $, $ 20^2 = 400 $, sum is 841, square root is 29. So $ CD = 29 $.
## Step3: Find $ m\angle ABE $
In triangle $ ABE $, we know $ \angle BAE = 51^\circ $, and since diagonals are perpendicular, $ \angle AEB = 90^\circ $. So $ m\angle ABE = 90^\circ - 51^\circ = 39^\circ $.
## Step4: Find $ m\angle BCE $
In triangle $ BCE $, $ \angle BEC = 90^\circ $ (diagonals are perpendicular), $ EC = 20 $, $ BE $ is some length, but wait, in triangle $ BCE $, since it's a right triangle, and we know that in a kite, the diagonal $ BD $ bisects the angles at $ B $ and $ D $. Wait, no, the diagonal that is the axis of symmetry bisects the angles. Wait, actually, in triangle $ BCE $, it's a right triangle, so $ m\angle BCE = 90^\circ $? No, wait, no. Wait, $ \angle BEC = 90^\circ $, so $ m\angle BCE = 90^\circ - m\angle CBE $. But we know that in triangle $ ABE $, $ m\angle ABE = 39^\circ $, and since $ AB = BC $ (kite sides), triangles $ ABE $ and $ CBE $ are congruent? Wait, no, $ AB = BC $, $ BE $ is common, $ AE = EC $, so triangles $ ABE $ and $ CBE $ are congruent (SSS). So $ \angle ABE = \angle CBE = 39^\circ $, so in triangle $ BCE $, $ m\angle BCE = 90^\circ - 39^\circ = 51^\circ $? Wait, no, wait, let's re-examine. Wait, in triangle $ ABE $, angles are $ 51^\circ $ (at $ A $), $ 90^\circ $ (at $ E $), so $ 39^\circ $ at $ B $. Then, since $ AB = BC $, $ AE = EC $, $ BE = BE $, so triangle $ ABE \cong \triangle CBE $ (SSS). Therefore, $ \angle BCE = \angle BAE = 51^\circ $? Wait, no, $ \angle BAE = 51^\circ $, $ \angle BCE $ would be equal to $ \angle BAE $? Wait, no, in triangle $ BCE $, angle at $ C $ is $ \angle BCE $, angle at $ B $ is $ \angle CBE = 39^\circ $, angle at $ E $ is $ 90^\circ $, so $ \angle BCE = 180 - 90 - 39 = 51^\circ $. Yes, that makes sense.

# Answers:
1. $ AC = 40 $
2. $ CD = 29 $
3. $ m\angle ABE = 39^\circ $
4. $ m\angle BCE = 51^\circ $

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QUESTION IMAGE

Question

Explanation:

Step1: Find AC

Step2: Find CD

Step3: Find \( m\angle ABE \)

Step4: Find \( m\angle BCE \)

Answer: